Find $\int_{0}^{2}(x^{2}+1) dx$ as the limit of a sum.

By definition,$\int_a^b f(x) dx = (b - a) \lim_{n \to \infty} \frac{1}{n} \sum_{r=0}^{n-1} f(a + rh)$,where $h = \frac{b-a}{n}$.
In this problem,$a = 0$,$b = 2$,$f(x) = x^2 + 1$,and $h = \frac{2-0}{n} = \frac{2}{n}$.
Therefore,$\int_0^2 (x^2 + 1) dx = 2 \lim_{n \to \infty} \frac{1}{n} \sum_{r=0}^{n-1} f(\frac{2r}{n})$.
$= 2 \lim_{n \to \infty} \frac{1}{n} \sum_{r=0}^{n-1} [(\frac{2r}{n})^2 + 1] = 2 \lim_{n \to \infty} \frac{1}{n} [\sum_{r=0}^{n-1} \frac{4r^2}{n^2} + \sum_{r=0}^{n-1} 1]$.
$= 2 \lim_{n \to \infty} [\frac{4}{n^3} \sum_{r=0}^{n-1} r^2 + \frac{1}{n} \sum_{r=0}^{n-1} 1]$.
Using $\sum_{r=0}^{n-1} r^2 = \frac{(n-1)n(2n-1)}{6}$ and $\sum_{r=0}^{n-1} 1 = n$,we get:
$= 2 \lim_{n \to \infty} [\frac{4}{n^3} \cdot \frac{(n-1)n(2n-1)}{6} + \frac{1}{n} \cdot n]$.
$= 2 \lim_{n \to \infty} [\frac{2}{3} \cdot \frac{(n-1)(2n-1)}{n^2} + 1] = 2 [\frac{2}{3} \cdot 2 + 1] = 2 [\frac{4}{3} + 1] = 2 [\frac{7}{3}] = \frac{14}{3}$.